User blog:Nordin.hammadi/The C power
As i defined E 1>x = (1>x)^(1>x) = 1>(x>x) I define the C power,C of Complex Extended Own Power,as C 1>x = (((1>x)^(1>x))^(1>x))^(1>x)......(1>x)times (1>x) in the power tower above base number(1>)...... From power math it can be derived that C 1>x = (1>x) ^E (1>x)= 1>(x>(x>x)) because from math we know (a^b)^c = a^(bc) so in the power tower of C 100 = C 1>2 we get 100x100x100.....100times100 in the sum.....=100^100 = E 100 and in the power tower of C 1>x we get (1>x)(1>x)(1>x)........(1>x)times..................= E 1>x and thus C 1>x = (1>x)^(E 1>x) and 1>x ^ ((1>x)^(1>x)) = (1>x)^(1>(x>x))= 10 to the power of (x((1>(x>x)))= 10 to the power of x>(x>x) or 1>(x>(x>x)) This is the mathematical proof,in the next part on this subject i will explore the patterns of C and E powers. A list demonstrates the patterns best as shown below; 1E 1>1 = 1>10 1C 1>1 = 1>(1>10) 1E 1>2 = 1>200 1C 1>2 = 1>(2>200) 1E 1>3 = 1>3000 1C 1>3 = 1>(3>3000) 1E 1>4 = 1>(4>4) 1C 1>4 = 1>(4>(4>4)) ..... 1E 1>9 = 1>(9>9) 1C 1>9 = 1>(9>(9>9)) 1E 1>10= 1C 1>10=1>(1>((1>10)~1)) 2E 1>1= 1>(1>11) 2C 1>1 = 1>(1>((1>((1>9)~8))~10)) 2E 1>2 =1>(2>202) 2C 1>2 = 1>(2>((2>((2>197)~197))~(2>2))) 2E 1>3 =1>(3>3003) 2C 1>3 = 1>(3>((3>((3>2996)~2996))~(3>3))) 2E 1>4 =1>(4>((4>3)~4)) 2C 1>4 = 1>(4>((4>((4>(3~(9re3)~5))~(3~(9re3)~5)))~(4>4))) ..... 2E 1>9 =1>(9>((9>8)~9)) 2E 1>10= 3E 1>1 = 1>(1>((1>9)~11)) 3E 1>2 = 1>(2>((2>199)~202)) 3E 1>3 = 1>(3>((3>2999)~3003)) 3E 1>4 = 1>(4>((4>(3~(9re4)))~((4>3)~4))) ...... 3E 1>9 = 1>(9>((9>(8~(9re9)))~((9>8)~9))) 3E 1>10 = 4E 1>1 = 1>(1>((1>(9re11))~(1>9)~11)) Now i will progress in E powers only with base number 1>1; 5E 1>1 = 1>(1>( (1>(9re((1>9)~11)))~ (1>(9re11)) ~(1>9) ~11)) 6E 1>1 = 1>(1>( (1>(9re((1>9re11)~(1>9)~11))) ~(1>(9re((1>9)~11)))~ (1>(9re11)) ~(1>9) ~11)) 7E 1>1 = 1>(1>( (1>(9re( (1>9re(1>9~11))~(1>9re11)~(1>9)~11))) ~repetition of the above)) 8E 1>1 = 1>(1>( (1>(9re( 1>9re((1>9re11)~(1>9)~11)~............... 9E 1>1 = 1>(1>( (1>(9re( 1>9re((1>9re(1>9~11))~(1>9re11)~1>9)~11)~......... etc. etc. So we could say for 1E 1>1 = 1>1 2E 1>1 = 1>(1>1) 3E 1>1 = 1>(1>( (1>9re0)~1) ) 4E 1>1 = 1>(1>( (1>9re1)~(1>9re0)~1) ) 5E 1>1 = 1>(1>( (1>9re((1>9re0)~1))~(1>9re1)~(1>9re0)~1) ) no matter how far we go,the outcom of the number will be in the form of 1>(1>( (1>(9re(z)))~(1>(9re(y)))~...................~(1>(9re(11)))~(1>(9re(1)))~(1>(9re(0)))~1 ) ) where z ofcourse is not necesarily the 26th in sequence but could be any last sequence,the 5th,the 24th,the millionth etc. When we aplly this listing on E powers of 1>2 base number the form of the number will only change in this way; 1>(2>( (2>(1~(9re(z))))~(2>(1~(9re(y))))~....................~(2>(1~(9re(202))))~(2>(1~(9re(2))))~(2>(1))~2 ) ) Note that the z to a valeus here will be different than with 1>1 as base number,that can be seen at the tails of the number forms. When we list the values from a to z say a is the last in sequence (the first from right on) from 4E on with base number 1>1 the list look as follows; a= (1>0)~1 b= (1>9re1)~(1>0)~1 c= (1>(9re(a)))~(1>9re1)~(1>0)~1 d= (1>(9re(b)))~(1>(9re(a)))~(1>9re1)~(1>0)~1 e= 1>(9re©))~(1>(9re(b)))~(1>(9re(a)))~(1>9re1)~(1>0)~1 ........... (n)th letter= (1>(9re((n-2)th letter)))~.....................~repetition of the above all down to ~(1>0)~1 The same principle goes for 1>2 or 1>3 or any base number. only then the last term in the sequence the (n)th letter for base number = 1>m will be ((m-1)>9re((n-2)th letter))). Category:Blog posts